prove that $H\cap K$ have finite index in G If $G$ is a group and $H,K$ are two subgroups of finite index in $G$, prove that $H\cap K$ is of finite index in $G$. Can you find upper bound for index of $H\cap K$  in $G$?
Since $a(H\cap K$) $\subseteq aH\cap aK$, and since there are only a finite many choices for $aH$ and $aK$, there are only finite many choices for $a(H\cap K$)
Is it correct ?
What about upper bound of index of $H\cap K$?
 A: If you are comfortable with group actions: Let G act on the set $S=G/H \times G/K$ (as coset spaces) by the law $g \star (aH,bK)=(gaH,gbK)$.Consider the element $(H,K)$ of $S$. Stab($(H,K)=H \cap K)$. Now we know that $[G: Stab((H,K)]=|Orb((H,K))| \leq |S|=[G:H][G:K]$. This upper bound is achievable (not too hard to find an example).
A: By multiplicativity of the subgroup index, we have
$$
[G: H \cap K] = [G : K] [K : H \cap K] \, .
$$
We claim that $[K : H \cap K] \leq [G : H]$.  Define a map
\begin{align*}
f: \frac{K}{H \cap K} &\to \frac{G}{H}\\
k (H \cap K) &\mapsto kH \, .
\end{align*}
It is straightforward to show that $f$ is well-defined.  We now show $f$ is injective.  Suppose $k_1 H = f(k_1(H \cap K)) = f(k_2(H \cap K)) = k_2 H$. Then $k_2^{-1} k_1 \in H$, so $k_2^{-1} k_1 \in H \cap K$, hence $k_1(H \cap K) = k_2(H \cap K)$ as desired.  Thus $f$ is injective, so $[K : H \cap K] \leq [G : H]$.  Therefore
$$
[G: H \cap K] = [G : K] [K : H \cap K] \leq [G : K] [G : H] \, .
$$
Since the indices on the righthand side are finite by assumption, this shows that the index of $H \cap K$ is also finite.
A: If $x \in aH \cap aK$, then $xa^{-1} \in H$ and $xa^{-1} \in K$. So what can you say about $x$ and $a$ modulo $H \cap K?$
A: I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is the same problem as Problem 12 on p.47 in Herstein's book.
Since I could not solve this problem, I referred to "An Introduction to Algebraic Systems" (in Japanese) by Kazuo Matsuzaka.
The following 3 problems are in this Matsuzaka's book:

Problem 2 (on p.60)
When $H$ is a subgroup of a group $G$, we denote the index of $H$ in $G$ by $i_G(H)$.
Suppose $H,K$ be subgroups of a group $G$ such that $H\supset K$.
Suppose $i_G(H)<+\infty$ and $i_H(K)<+\infty$.
Then, $i_G(K)<+\infty$ and $$i_G(K)=i_G(H)i_H(K)$$ holds.
My solution:
Let $i_G(H)=m$ and $i_H(K)=n$.
Then, we can write $G/H=\{g_1H,\dots,g_mH\}$ for some $g_1,\dots,g_m\in G$.
Then, we can write $H/K=\{h_1K,\dots,h_nK\}$ for some $h_1,\dots,h_n\in H$.
Let $x$ be an arbitrary element of $G$.
Then, $x\in g_iH$ for some $i\in\{1,\dots,m\}$.
So, $x=g_ih$ for some $h\in H$.
$h\in h_jK$ for some $j\in\{1,\dots,n\}$.
So, $h=h_jk$ for some $k\in K$.
Therefore, $x=g_ih_jk\in g_ih_jK$.
So, $i_G(K)\leq m\cdot n$.
Next we prove that $\#\{g_ih_jK\mid i\in\{1,\dots,m\}\text{ and }j\in\{1,\dots,n\}\}=m\cdot n$.
Let $(i,j)\neq (k,l)$, where $i,k\in\{1,\dots,m\}$ and $j,l\in\{1,\dots,n\}$.
We prove that $g_ih_jK\neq g_kh_lK$.
Assume that $g_ih_jK\neq g_kh_lK$ doesn't hold.
Then, $g_ih_jK=g_kh_lK$ must hold.
So, $g_ih_j=g_kh_lk$ for some $k\in K$.
$g_i=g_kh_lkh_{j}^{-1}\in g_kH$.
So, $g_iH=g_kH$.
So, $i=k$.
Since $g_ih_jK=g_ih_lK$, $h_jK=h_lK$.
So, $j=l$.
So, $(i,j)=(k,l)$.
This is a contradiction.
So, $g_ih_jK\neq g_kh_lK$ holds.
So, $\#\{g_ih_jK\mid i\in\{1,\dots,m\}\text{ and }j\in\{1,\dots,n\}\}=m\cdot n$ holds.
Therefore, $i_G(K)=i_G(H)i_H(K)$.


Problem 3 (on p.60)
Suppose $H,K$ be subgroups of $G$ and $i_G(K)<+\infty$.
Then, $i_H(H\cap K)<+\infty$ and $i_H(H\cap K)\leq i_G(K)$ holds.
My solution:
$H/H\cap K\ni h(H\cap K)\mapsto hK\in G/K$ is well-defined since $h^{-1}h^{'}\in H\cap K$ implies $h^{-1}h^{'}\in K$.
Suppose that $hK=h^{'}K$.
Then $h^{-1}h^{'}\in K$.
Since $h,h^{'}\in H$, $h^{-1}h^{'}\in H\cap K$.
So, $h(H\cap K)=h^{'}(H\cap K)$.
So, this well-defined mapping is injective.
So, $i_H(H\cap K)<+\infty$ and $i_H(H\cap K)\leq i_G(K)$ must hold.


Problem 4 (Poincare) (on p.60)
Suppose that $H_1,\dots,H_n$ be subgroups of $G$ and $i_G(H_i)<+\infty$ for all $i\in\{1,\dots,n\}$.
Then, $i_G(H_1\cap\dots\cap H_n)<+\infty$ and $$i_G(H_1\cap\dots\cap H_n)\leq i_G(H_1)\cdots i_G(H_n)$$ hold.
My solution:
Obviously, $i_G(H_1)<+\infty$ and $i_G(H_1)\leq i_G(H_1)$ hold.
Assume that $i_G(H_1\cap\dots\cap H_k)<+\infty$ and $i_G(H_1\cap\dots\cap H_k)\leq i_G(H_1)\cdots i_G(H_k)$ hold.
By Problem 3, $$i_{H_{k+1}}(H_1\cap\dots\cap H_k\cap H_{k+1})<+\infty$$ and $$i_{H_{k+1}}(H_1\cap\dots\cap H_k\cap H_{k+1})\leq i_G(H_1\cap\dots\cap H_k)$$ hold.
By Problem 2, $$i_{G}(H_1\cap\dots\cap H_k\cap H_{k+1})=i_G(H_{k+1})i_{H_{k+1}}(H_1\cap\dots\cap H_k\cap H_{k+1})<+\infty$$ holds.
So, $$i_{G}(H_1\cap\dots\cap H_k\cap H_{k+1})\leq i_G(H_{k+1})i_G(H_1\cap\dots\cap H_k)\leq\\ i_G(H_1)\cdots i_G(H_k)i_G(H_{k+1})$$ holds.

